Math Forum

by **elim** » Mon Mar 08, 2010 9:31 am

Determine the convergence of $\displaystyle{\lim_{x \to \infty} \frac{x}{\ln x} \sum_{n=0}^{\infty} \frac{1}{a^n+x}} \quad$ where $a>1$

by **jason.spade** » Mon Mar 08, 2010 12:31 pm

I like this question very much, and I had to think about it for a few moments. The key idea for my solution was to recall that the limit of the sum is the sum of the limits. I actually tried two other things before this method struck me, and there's still a little uncertainty. But it can be handled.

How is that for a start?

by **elim** » Mon Mar 08, 2010 2:41 pm

I found this problem a few months ago. Still don't know how to solve it. But I know how to find the limit if the limit does exist.
For this problem, exchange the order of limit and summation is not allowed unless you can prove that you can do it. That makes the thing hard isn't it?

by **dman315** » Mon Mar 08, 2010 4:31 pm

use the dominated convergence theorem to switch the limits.

by **jason.spade** » Mon Mar 08, 2010 4:34 pm

Forgive me, but why can't we interchange the order in this question? I admit I haven't really had to deal with analysis for a while, but I have studied it. I seem to recall a theorem which allows the interchange of the order if the terms are dominated. Perhaps I should review.

by **jason.spade** » Mon Mar 08, 2010 4:49 pm

Oh, I now see that dman beat me to the chase.

by **elim** » Mon Mar 08, 2010 8:13 pm

If $\lim_{x \to \infty} h(x) = A$ exists, then for any $\left\{ x_n\right\}$ with $\lim_{n \to \infty} x_n = \infty$ , we have $\lim_{n\to \infty} h(x_n) = A$

Now let $f(x) = x \sum_{n=0}^{\infty} \frac{1}{a^n+x}, \quad g(x) = \ln x$

Then we have $\frac{f(a^{k+1})-f(a^{k})}{g(a^{k+1})-g(a^{k})}=\frac{a^{k+1}\sum_{n=0}^{\infty}\frac{1}{a^n+a^{k+1}}-a^{k}\sum_{n=0}^{\infty}\frac{1}{a^n+a^{k}}}{\ln a} =$
$=\frac{\sum_{n=0}^{\infty}\frac{1}{a^{n-k-1}+1}-\sum_{n=0}^{\infty}\frac{1}{a^{n-k}+1}}{\ln a}=\frac{1}{(a^{-k-1}+1)\ln a} \to \frac{1}{\ln a} \quad (k \to \infty)$

By Stolz theorem, $\lim_{n \to \infty} \frac{f(a^n)}{g(a^n)} = \frac{1}{\ln a}$

Therefore the original limit necessarily equals to $\frac{1}{\ln a}$ provided it exists
But again, does the limit exist?

by **elim** » Tue Mar 09, 2010 6:54 pm

dman315 wrote:use the dominated convergence theorem to switch the limits.

Could you please give some more details? Thanks

by **dman315** » Wed Mar 10, 2010 10:01 am

find a dominating function that is summable, then you can interchange the orders. In this case there is a simple dominating function.

by **elim** » Wed Mar 10, 2010 10:50 am

dman315 wrote:find a dominating function that is summable, then you can interchange the orders. In this case there is a simple dominating function.

Once the order been changed, you got 0 valued limit but this contradicts my necessary condition $\frac{1}{\ln a}$
My guess is that a summable dominating function is not enough, need that function's limit be finite as x goes to infinity.

by **dman315** » Wed Mar 10, 2010 2:47 pm

I looked at it too quickly, there is no dominating function that is why you can't interchange the orders.

by **elim** » Wed Mar 10, 2010 3:21 pm

Yes. That makes the problem tough. I have some estimation about $\lim sup$ and $\lim inf$ , they are both finite and positive. But I wasn't accurate enough to be able to tell the convergence.

by **mattpi** » Thu Mar 11, 2010 6:28 pm

Consider $f_a(x)=\frac{x}{\ln x}\sum_{n=0}^\infty\frac1{a^n+x}.$ Note that this can be written as $f_a(x)=\frac{x}{\ln x}\int_0^\infty\frac1{a^{\lfloor t\rfloor}+x}dt,$ and so

$\frac{x}{\ln x}\int_0^\infty\frac1{a^t+x}dt\ <\ f_a(x)\ <\ \frac{x}{\ln x}\int_0^\infty\frac1{a^{t-1}+x}dt.$

Note that the right-hand integral can be rewritten as $\int_{-1}^\infty\frac1{a^t+x}dt.$ We can evaluate these integrals as follows:

$\begin{align} \int_0^\infty\frac1{a^t+x}dt&=\int_0^\infty\frac{a^{-t}}{1+xa^{-t}}dt\\ &=\frac1{\ln a}\int_0^1\frac1{1+xu}du,\qquad\text{using the substitution }u=a^{-t},\ du=-a^{-t}\ln a\, dt\\ &=\frac1{\ln a}\left[\frac1x\ln(1+xu)\right]_{u=0}^{u=1}\\ &=\frac{\ln(1+x)}{x\ln a}. \end{align}$

Similarly, $\int_{-1}^\infty\frac1{a^t+x}dt=\frac1{\ln a}\int_0^a\frac1{1+xu}du=\frac{\ln(1+ax)}{x\ln a}=\frac{\ln(a^{-1}+x)+\ln a}{x\ln a}.$

Therefore, $\frac{\ln(1+x)}{\ln x\, \ln a}\,<\,f_a(x)\,<\,\frac{\ln(a^{-1}+x)}{\ln x\,\ln a}+\frac1{\ln x}.$ As we take $x\to\infty,$ the expressions on the left and right both converge to $1/\ln a,$ thereby establishing the convergence of $f_a(x)$ to this quantity.

by **elim** » Thu Mar 11, 2010 10:58 pm

Great! Thanks mattpi !

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